1. 已知a≠2，求証:4a/(4+a^2)<1。

1)1 - 4a / (4+a^2) = (a^2 - 4a + 4) / (4+a^2) = (a - 2)^2 / (4+a^2) 因 (a - 2)^2 >= 0 , (4 + a^2) > 0又 a≠2 故 (a - 2)^2 > 0 ,故(a - 2)^2 / (4+a^2) > 01 - 4a / (4+a^2) > 04a/(4+a^2) < 1
2)(a^2+1)/2a - 1= (a^2 - 2a + 1) / 2a= (a - 1)^2 / 2a >= 0(a^2+1)/2a - 1 >= 0(a^2+1)/2a ≥ 1
3)60 + 28 =< a + b =< 84 + 3388 =< a + b =< 117
60 - 33 =< a - b =< 84 - 2827 =< a - b =< 56
60/33 =< a/b =< 84/2820/11 =< a/b =< 3

2010-09-09 23:34:45 補充：
更正 : 沒 = 的,

3)

60 + 28 < a + b < 84 + 33
88 < a + b < 117

60 - 33 < a - b < 84 - 28
27 < a - b < 56

60/33 < a/b < 84/28
20/11 < a/b < 3

1. Since square of any real number is positive. (a - 2)^2 > 0 on condition a is not equal to 2. So a^ - 4a + 4 > 0
a^ + 4 > 4a
1 > 4a/(a^2 + 4) [Note that (a^2 + 4) is always positive, thus not affecting the inequality sign.]
2. Similarly, to Q1. (a - 1)^2 > or equal to 0. [equal to 0 when a = 1].
So a^2 - 2a + 1 > or equal to 0
a^2 + 1 > or equal to 2a
(a^2 + 1)/2a > or equal to 1 [Note a > 0, so dividing by 2a does not affect the inequality sign.]
3. Refer to other source.